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Prove that there are no two functions $f$ and $g$ defined over real numbers that satisfy either of the following functional equations: $f(x)g(y) = x+y$ and $f(x) + g(y) = xy$.

I think we should prove this using casework. That is, show it is impossible for the first equation and then show it is impossible for the second. I saw that for the first equation we get $f(x)g(x) = 2x$ and for the second $f(x) +g(x)=x^2$. I am not sure if those will help or not.

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    $\begingroup$ What else would you want to do except treat the cases separately, since they are completely different ? This being said, try to look at what happens for $x=0$ and $y=0$. $\endgroup$ – Captain Lama Apr 9 '16 at 19:29
  • $\begingroup$ @CaptainLama Yes, I thought about that. Then we just have $f(0)g(0) = 0$ and $f(0)+g(0) = 0$. $\endgroup$ – user19405892 Apr 9 '16 at 19:30
  • $\begingroup$ You're looking at $x=y=0$. But you don't have to take them both $0$ at the same time. $\endgroup$ – Captain Lama Apr 9 '16 at 19:37
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Your idea is not wrong, but it's probably not the simplest way to go.

For the first, let's start by looking at $y=0$, that gives us: $$ f(x)g(0)=x $$ or (if $g(0)\neq 0$): $$ f(x)=\frac{x}{g(0)} $$

If we look at what happens when $x=0$, we similarily get: $$ g(y)=\frac{y}{f(0)} $$

We can substitute both of those expressions into the equation we wanted, and get: $$ f(x)g(y)=\frac{xy}{f(0)g(0)} $$ That's not $x+y$.

The other one can be done in the same way.

$x=0$ and $y=0$ are curves in $\mathbb R^2$, and so is $x=y$ that you substituted, and the "trick" is just to use what happens on two different curves to obtain expressions that can be used to arrive at a contradiction, so rewriting what you had found and finding another curve you could probably have gotten through, but in these cases the arguments get a lot simpler when you consider curves where one of the variables is constant (I'm sure cases can be found where $x=0$ and $y=0$ are not good choices)

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For the first,

Assume that we can find $f,g$ such that $f(x)g(y) = x + y$. Note that, $f(x)g(0) = x \;\; \forall x$. This implies that $g(0) \neq 0$. Similarly, $f(0)g(y) = y \;\; \forall y$. This implies that $f(0) \neq 0$. However, $f(0)g(0) = 0$. This implies that either $f(0) = 0$ or $g(0) = 0$. Contradiction.

For the second,

Assume that we can find $f,g$ such that $f(x) + g(y) = xy$. Note that, for $x = 0$, $g(y) = -f(0) \;\; \forall y$. Similarly, for $x = 1$, $g(y) = y - f(1) \;\; \forall y$. Contradiction.

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